# Do semi-continuous functions generate bounded Borel measurable functions as a $C^*$-algebra?

This question is related to Question 2 of my previous posting.

Question. Let $\mu$ be a Radon measure on a compact Hausdorff space $\Omega$ and $L^{\infty}(\Omega,\mu)$ the set of essentially bounded Borel measurable functions on $\Omega$. Suppose that $S$ is the set of bounded lower or upper semi-continuous functions on $\Omega$. Does $S$ generate $L^{\infty}(\Omega,\mu)$ as a $C^*$-algebra?

It suffices to consider whether the indicator function of any Borel set is obtained from $S$ by algebraic operations and (essential-supremum) norm limits. If necessary, you may assume $\Omega$ to be second countable. Thank you.

No. The bounded Baire class one functions on $[0,1]$ are stable under uniform limits and hence constitute a C*-algebra. This C*-algebra contains every semicontinuous function on [0,1]. Every function in $L^\infty[0,1]$ is equal almost everywhere to a Baire class two function, but not a Baire class one function. (The previous version of my answer neglected this essential point.)
Any upper or lower semicontinuous function is continuous almost everywhere in the sense of Baire category (since it is a pointwise limit of a sequence of continuous functions, at least when $\Omega$ is compact metrizable). The algebra of Baire-a.e. continuous functions is itself a $C^\ast$-algebra. So the answer to your question is no, once we show that there exists a function $f \in L^\infty$ that is not $\mu$-equivalent to a Baire-a.e. continuous function. For an example we may take any indicator function of a set $S$ such that both $\mathrm{supp}\ \mu\restriction S$ and $\mathrm{supp}\ \mu\restriction (\Omega \setminus S)$ equal $\Omega$.